# Basic proof of units in a ring with identity.

1. Feb 28, 2013

### shamus390

1. The problem statement, all variables and given/known data

Let R be a ring with identity, and a,b are elements in R. If ab is a unit, and neither a nor b is a zero divisor, prove a and b are units.

2. Relevant equations

If ab is a unit then (ab)c=1=c(ab) for some c in R.

3. The attempt at a solution
Assume both a and b are not zero divisors, and denote the identity element of R as 1.

Since ab is a unit in R, there exists some c such that (ab)c = 1.
(ab)c=1 $\Rightarrow$ (ca)b = 1 $\Rightarrow$ b is a unit.

Similarly, (ab)c= 1 $\Rightarrow$ a(bc) $\Rightarrow$ a is a unit.

Does the above sufficiently prove the claim in the problem statement? I feel like I'm missing something.

Thanks!

2. Feb 28, 2013

### Dick

Yes, you should feel uncomfortable with that. You didn't use that a and b weren't zero divisors. If (ab)c=1 then you can regroup that to a(bc)=1. That doesn't prove a is a unit. You also need to show (bc)a=1. Don't assume the ring is commutative.

3. Mar 1, 2013

### shamus390

Ah thank you, that was a foolish oversight, they wouldn't stipulate a and b weren't zero divisors if it wasn't necessary...

So if we let a, b and c be elements of R such that ab = ac, then:

ab-ac=0R = a(b-c) = 0R, and
a is not a zero divisor $\Rightarrow$ (b - c) = 0R $\Rightarrow$ (b-c) = 0R $\Rightarrow$ b = c

Does that allow me to claim the following?

Multiplying a(bc) = 1 by a to give a(bc)a = 1a , (1a=a1 by definition) then, a(bc)a = a1 and by cancellation, (bc)a = 1 $\Rightarrow$ a is a unit.

4. Mar 1, 2013

### Dick

Yes, that's the idea.